Deduce the expression for the electrostatic energy stored in a capacitor of capacitance 'C' and having charge 'Q'.
How will the (i) energy stored and (ii) the electric field inside capacitor be affected when it is completely filled with a dielectric material of dielectric constant 'K'?

Explanation

Parallel plate capacitor : Consider a parallel plate capacitor having two plane metallic plates A and B, placed parallel to each other. The plates carry equal and opposite charges + Q and -Q respectively.
Suppose A be the area of each plate, 'd' the separation between the plates, k the dielectric constant of medium between the plates. If s is the magnitude of charge density of plates, then `sigma=(Q)/(A)`
Electric field, `E=(sigma)/(K epsilon_(0))" ...(i)"`
`in_(0)rarr` permittivity of free space.
The potential difference between the plates,
`V_(A)=ind=(sigma d)/(K in_(0))" ....(ii)"`
Putting the value of `sigma` we get
`V_(AB)=((Q//A)d)/(K in_(0))=(Qd)/(K in_(0)A)`
`therefore" Capacitance of capacitor,"`
`C=(Q)/(V_(AB))=(Q)/((Qd//K in_(0)A))`
`C=(K in_(0)A)/(d).`
This is a general expression for capacitance of parallel plate capacitor. Obviously, the capacitance is directly proportional to the dielectric constant of medium between the plates.